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irga5000 [103]
3 years ago
7

Consider the following reaction:

Chemistry
1 answer:
vfiekz [6]3 years ago
3 0

Answer:

Kc = 4.76

Explanation:

To find the concentrations of CO and H₂ at equilibrium, you have to set up an ICE (Initial, Change, Equilibrium) table.

    CO (g) + 2H₂(g) ⇌ CH₃OH (g)

I   0.32 M   0.53 M          0

C    -x            -2x            +x

E  0.32-x    0.53-2x        0.16 M

Since you know the concentration of CH₃OH at equilibrium, it would be equal to x since 0 + x = 0.16. So,

[CH₃OH] = 0.16 M

[CO] = 0.32 - 0. 16 = 0.16 M

[H₂] = 0.53 - 2(0.16) = 0.21 M

Now that you have all the concentrations at equilibrium, you can calculate the equilibrium constant.

Kc  = products ÷ reactants

     = [CH₃OH] ÷ [CO][H₂]²

     =  0.16 ÷ (0.16)(0.21)

Kc =   4.76

The equilibrium constant at this temperature is 4.76.

Hope this helps.

     

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Beaker A contains 2.06 mol of copper ,and Barker B contains 222 grams of silver.Which beaker the larger number of atom?
Dmitry [639]

Answer:

The number of copper atoms 12.405 ×10²³ atoms.  

The number of silver atoms  13.13 ×10²³ atoms.

Beaker B have large number of atoms.

Explanation:

Given data:

In beaker A

Number of moles of copper = 2.06 mol

Number of atoms of copper = ?

In beaker B

Mass of silver = 222 g

Number of atoms of silver = ?

Solution:

For beaker A.

we will solve this problem by using Avogadro number.

The number 6.022×10²³ is called Avogadro number and it is the number of atoms in one mole of substance.

While we have to find the copper atoms in 2.06 moles.

So,

63.546 g = 1 mole = 6.022×10²³ atoms

For 2.06 moles.

2.06 × 6.022×10²³ atoms

The number of copper atoms 12.405 ×10²³ atoms.  

For beaker B:

107.87 g = 1 mole = 6.022×10²³ atoms

For 222 g

222 g / 101.87 g/mol = 2.18 moles

2.18 mol × 6.022×10²³ atoms = 13.13 ×10²³ atoms

8 0
3 years ago
I don’t understand this question and need help!!
Dafna1 [17]
Medium about 3 second? Not sure lol just need more points honestly lol
6 0
2 years ago
Calculate the enthalpy of formation of butane, C4H10, using the balanced chemical
melisa1 [442]

Answer:

-125.4

Explanation:

Target equation is 4C(s) + 5H2(g) = C4H10

These are the data equations for enthalpy of combustion

  1. C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
  2. H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
  3. 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse

To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...

Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.

7 0
3 years ago
What is the total pressure of a container in kpa if the partial pressures are 1.25 atm and 66.7 psi?
STatiana [176]

Answer:

586 kpa(kilopascal/1000 pascals)

Explanation:

given 1.24 atm(standard atmosphere), and 66.7 psi(pound force per square inch).

To find the total pressure we should use dalton's law of partial pressures which is the sum of the pressures of each individual gas.

then we convert them to pascals and divide by 1000 to get the measurement in kilopascal.

knowing that 1 atmosphere is proportional to around 14.696 psi. We can multiply our given measure of atm by that and sum it by psi like so. 1.24×14.6959 = 18.22298.

Then,

18.22298+ 66.7 = 84.92298

psi.

Since 1 psi is proportional to around 6894.76 pascals. 1 psi will be 68.9476 kilopascal. 84.92298 * 6.89476 = 585.523336 ≈ 586

4 0
3 years ago
What is the volume of a rock that has a mass of 50 grams and a density of<br> 5 g/ml?
Brut [27]

Explanation:

density = mass/ volume

so,

volume = mass/ density

volume = 50 g/ 5 g/ml³

volume = 10 ml³

3 0
3 years ago
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